Question: $h(n) = -6n+2-3(f(n))$ $g(x) = -x^{3}+x^{2}$ $f(n) = -3n+5+g(n)$ $ h(f(2)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(2)$ . Then we'll know what to plug into the outer function. $f(2) = (-3)(2)+5+g(2)$ To solve for the value of $f$ , we need to solve for the value of $g(2)$ $g(2) = -2^{3}+2^{2}$ $g(2) = -4$ That means $f(2) = (-3)(2)+5-4$ $f(2) = -5$ Now we know that $f(2) = -5$ . Let's solve for $h(f(2))$ , which is $h(-5)$ $h(-5) = (-6)(-5)+2-3(f(-5))$ To solve for the value of $h$ , we need to solve for the value of $f(-5)$ $f(-5) = (-3)(-5)+5+g(-5)$ To solve for the value of $f$ , we need to solve for the value of $g(-5)$ $g(-5) = -(-5)^{3}+(-5)^{2}$ $g(-5) = 150$ That means $f(-5) = (-3)(-5)+5+150$ $f(-5) = 170$ That means $h(-5) = (-6)(-5)+2+(-3)(170)$ $h(-5) = -478$